OCR GCSE Chemistry A: Bonding (J248)

This study guide teaches sub-topic C2.2 Bonding (statements C2.2a to C2.2i, plus the mathematical learning outcomes CM2.2i to CM2.2iii) from Topic C2 of the OCR GCSE (9–1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026), for first assessment in 2018. None of the C2.2 statements is Higher Tier only, so all of it is assessed on both tiers. The specification says Paper 1 (Foundation) and Paper 3 (Higher) assess Topics C1–C3 and C7, and Papers 2 and 4 assume knowledge of C1–C3.

Use it with the Bonding revision notes and the Bonding practice questions. For the full course, see the OCR GCSE Chemistry hub and the printable J248 checklist. To find your weak spots first, take a free 10-minute diagnostic.

What this sub-topic covers

Ref	You should be able to
C2.2a	Describe metals and non-metals; explain their differences using physical and chemical properties (including ion formation and oxides)
C2.2b	Link the atomic structure of metals and non-metals to where they sit in the Periodic Table
C2.2c	Link group number and period number to electron arrangement and atomic number
C2.2d	Describe and compare bonding in ionic compounds, simple molecules, giant covalent structures, polymers and metals
C2.2e	Explain bonding using electrostatic forces and the transfer or sharing of electrons
C2.2f	Construct dot and cross diagrams for simple covalent and binary ionic substances
C2.2g	Describe the limitations of dot and cross diagrams, ball and stick models, and 2D and 3D drawings
C2.2h	Explain how the reactions of elements depend on electron arrangement and atomic number
C2.2i	Explain how Mendeleev’s table was refined into the modern table using atomic number
CM2.2i–iii	Estimate the size and scale of atoms and nanoparticles; switch between 3D and 2D representations; translate between diagrams and numbers

The practical suggestion for this sub-topic is building ball and stick models of molecules. This page assumes you already know protons, neutrons and electrons from C1.2 Atomic structure.

Metals and non-metals (C2.2a)

Property	Typical metal	Typical non-metal
Appearance	Shiny when freshly cut	Dull (as solids)
Conducts electricity and heat	Yes, well	Poorly (graphite is the exception)
Melting and boiling point	Mostly high	Mostly low; many are gases at room temperature
Solid behaviour	Malleable and ductile	Brittle
Ions formed	Positive ions (cations)	Negative ions (anions), or they share electrons
Oxide	Basic: reacts with acids; soluble ones give alkaline solutions	Acidic: soluble ones give acidic solutions

Both react with oxygen to form oxides. Burning calcium gives calcium oxide (2Ca + O₂ → 2CaO). Burning sulfur gives sulfur dioxide (S + O₂ → SO₂). Calcium oxide neutralises acids. Sulfur dioxide dissolves in water to give an acidic solution.

Watch for exceptions: mercury is a liquid metal, and diamond (a non-metal) has a very high melting point.

Atomic structure and position in the table (C2.2b, C2.2c)

Electrons fill shells in order. Up to calcium (Z = 20), shell 1 takes a maximum of 2 electrons and shells 2 and 3 take a maximum of 8 each. Two rules link the arrangement to the table:

• Group number (1 to 7) tells you how many electrons sit in the outermost shell; in group 0 that shell is full.
• Period number = the number of occupied shells.

An uncharged atom has as many electrons as protons, so its atomic number fixes the electron arrangement, and the arrangement fixes the position.

Worked example 1. Calcium has atomic number 20. Give its electron arrangement and its position.

• 20 electrons: 2 in shell 1, 8 in shell 2, 8 in shell 3, 2 in shell 4. Arrangement 2,8,8,2.
• Four occupied shells, so period 4. Two outer electrons, so group 2.

Metals sit on the left and in the centre; their atoms have few outer electrons (usually 1 to 3) and lose them. Non-metals sit on the right; their atoms have 4 to 7 outer electrons and gain or share electrons. Group 0 atoms already have full outer shells.

Size and scale (CM2.2i)

Atomic radii and bond lengths are of the order of 10⁻¹⁰ m (0.1 nm). Nanoparticles are much bigger than atoms but still tiny: roughly 1 to 100 nm across.

Worked example 2. A silver nanoparticle is 14.5 nm across. Take the diameter of a silver atom as 0.29 nm. Estimate how many atoms span the particle.

• Number of atoms = 14.5 ÷ 0.29 = 50 atoms across.

Put both lengths in the same unit before dividing (1 nm = 1 × 10⁻⁹ m).

Why atoms bond: electrostatic forces (C2.2e)

A chemical bond is not a physical object. It is a strong electrostatic attraction between opposite charges.

• In ionic bonding, electrons transfer from metal atoms to non-metal atoms. The positive and negative ions attract each other.
• In covalent bonding, two non-metal atoms share a pair of electrons. Both nuclei (positive) attract the shared pair (negative), and that holds the atoms together.
• In metallic bonding, positive metal ions attract a sea of delocalised electrons.

When an atom becomes an ion, only the number of electrons changes. The nucleus, and so the number of protons, stays the same. A calcium ion, Ca²⁺, still has 20 protons but now has 18 electrons (2,8,8).

The five bonding types compared (C2.2d)

Structure	Particles	Bonds and arrangement
Ionic compound	Positive and negative ions	Ionic bonds act in every direction through a giant lattice of alternating ions
Simple molecule	Small molecules	Strong covalent bonds inside each molecule; only weak intermolecular forces between molecules
Giant covalent	Atoms	Atoms joined by covalent bonds in a continuous network: 3D in diamond and silicon dioxide, flat layers in graphite
Polymer	Very long molecules	Covalent bonds join repeating units into long chains; weaker forces between the chains
Metal	Positive ions and delocalised electrons	Regular layers of ions held by attraction to the shared electron sea

A sodium chloride crystal contains no molecules: each ion is surrounded by oppositely charged ions, and the formula gives only their ratio. Each poly(propene) chain holds thousands of C–C and C–H covalent bonds.

Turning a diagram into numbers (CM2.2ii, CM2.2iii)

Worked example 3. A ball and stick model of silicon dioxide shows each silicon atom joined to 4 oxygen atoms, and each oxygen atom joined to 2 silicon atoms. Deduce the formula.

• Count bonds from both ends. For n silicon atoms there are 4n Si–O bonds.
• Each oxygen uses 2 of those bonds, so the number of oxygen atoms = 4n ÷ 2 = 2n.
• Ratio Si : O = 1 : 2, so the formula is SiO₂.

You should also be able to go the other way: build a 3D model from a 2D drawing, and draw a 3D model flat.

Dot and cross diagrams (C2.2f)

Use dots for electrons from one atom and crosses for electrons from the other. At GCSE level you normally show outer shells only, unless the question asks for every shell.

Binary ionic compounds

Worked example 4. Draw a dot and cross diagram for calcium fluoride.

1. Calcium is 2,8,8,2. It loses 2 electrons to become Ca²⁺ (2,8,8).
2. Fluorine is 2,7. Each atom gains 1 electron to become F⁻ (2,8).
3. One calcium atom gives one electron to each of two fluorine atoms, so the formula is CaF₂.
4. Draw Ca²⁺ in square brackets with no electrons left in its old outer shell (or show the full 2,8,8 if you draw every shell). Put the charge outside the bracket at top right.
5. Draw two F⁻ ions, each in brackets with charge −. Each shows 7 dots and 1 cross in its outer shell (8 in total).

Charges must balance: (+2) + 2 × (−1) = 0. Always check this.

Simple covalent molecules

Worked example 5. Draw a dot and cross diagram for silicon tetrachloride, SiCl₄.

1. Silicon has 4 outer electrons. Each chlorine has 7 and is one short of 8.
2. Silicon pairs each of its 4 electrons (dots) with one electron (cross) from a different chlorine atom, giving four shared pairs in four overlap regions.
3. Each chlorine keeps its other 6 electrons as 3 lone pairs. Silicon has no lone pairs.
4. Check: Si has 4 × 2 = 8; each Cl has 2 + 6 = 8. Total outer electrons 4 + 4 × 7 = 32, which is 16 pairs (4 shared and 12 lone).

Worked example 6. Draw hydrogen cyanide, HCN.

1. Carbon (4 outer) shares one pair with H and three pairs with N.
2. Nitrogen (5 outer) puts 3 into the triple bond and keeps 1 lone pair.
3. Check: C has 8, N has 8, H has 2. Total outer electrons 1 + 4 + 5 = 10 = 4 shared pairs + 1 lone pair.

Limitations of models (C2.2g)

• Dot and cross diagrams show where electrons came from. Real electrons are identical, though, so the dot/cross split is only bookkeeping. They are flat, so they do not show the 3D shape, and they show shells as fixed rings.
• Ball and stick models show 3D shape and bond angles. But bonds are not sticks, atoms are not hard coloured spheres, and the gaps between atoms are exaggerated. For an ionic lattice, the sticks wrongly suggest that each ion is bonded to particular neighbours.
• 2D drawings (displayed formulae) show which atoms are joined, but not the shape. 3D drawings use wedges and dashes to suggest depth, but they are still projections onto paper.

Reactions and electron arrangement (C2.2h)

Members of one group share an outer-electron count, which is why their reactions look alike.

• Group 1 atoms each lose one electron. Each step down adds a shell, so the single outer electron sits further out with more inner electrons between it and the nucleus. The pull on it weakens, it leaves more readily, and reactivity rises down the group.
• Group 7 atoms each gain one electron. Going down, the incoming electron is further from the nucleus and attracted less strongly, so reactivity decreases down the group.
• Group 0 atoms already have full outer shells, so they are very unreactive.

Atomic number links all of this together: it sets the electron arrangement, which sets the group, which predicts the reactions.

Mendeleev to the modern table (C2.2i)

Mendeleev (1869) ordered elements by atomic weight and grouped elements with similar properties. He left gaps for elements not yet found and predicted their properties. Where atomic weight order put an element in the wrong family, he swapped the pair to keep the properties lined up.

Later work showed that atomic number (proton number) is the true ordering principle. Moseley’s X-ray measurements around 1913 gave each element an atomic number. Ordering by atomic number keeps the swapped pairs in the right families without any special exceptions.

Worked example 7. On average, argon atoms are slightly heavier than potassium atoms. Explain why argon still comes before potassium in the modern table.

• In mass order potassium would come first and sit with the noble gases, which does not suit a reactive metal.
• Atomic numbers are argon 18 and potassium 19. Ordering by atomic number puts argon in group 0 and potassium in group 1, matching their chemistry.

Common errors

• Writing that a positive ion has gained protons. It has lost electrons; the nucleus is unchanged.
• Calling sodium chloride a “molecule” or saying it has “intermolecular forces”.
• Leaving out brackets and charges on ionic dot and cross diagrams, or showing ion charges that do not balance.
• Drawing a lone pair inside a shared region.
• Claiming that melting a simple molecular solid splits its molecules into atoms; it is the weak attractions between molecules that are overcome.

Next steps

After this guide, go to the revision notes and then the practice set (both linked in the opening section). The purity and separating mixtures guide covers C2.1, and the J248 exam preparation page covers the whole course. The diagnostics show what still needs work.

Official syllabus

OCR GCSE (9–1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026), for first assessment in 2018, published by OCR. Sub-topic C2.2 Bonding.